The dimensions of force are:
[M L T-2]
Understanding the Dimensions of Force
Dimensional analysis is a powerful tool in physics that helps us understand the relationship between different physical quantities by identifying their fundamental dimensions: mass (M), length (L), and time (T). The dimension of a physical quantity tells us how it depends on these fundamental dimensions.
The question asks for the dimensions of force. To find this, we can use a fundamental equation involving force, such as Newton's second law of motion.
Newton's second law states that force ($\vec{F}$) is equal to the product of mass ($m$) and acceleration ($\vec{a}$).
Mathematically, this is expressed as: $$\vec{F} = m \vec{a}$$
To find the dimensions of force, we need the dimensions of mass and acceleration.
- The dimension of mass is simply [M].
- Acceleration is the rate of change of velocity, and velocity is the rate of change of displacement (length).
Let's find the dimensions of acceleration step-by-step:
- Dimension of displacement (length) = [L]
- Dimension of time = [T]
- Dimension of velocity = $\frac{\text{Dimension of Displacement}}{\text{Dimension of Time}} = \frac{[L]}{[T]} = [L T^{-1}]$
- Dimension of acceleration = $\frac{\text{Dimension of Velocity}}{\text{Dimension of Time}} = \frac{[L T^{-1}]}{[T]} = [L T^{-1} T^{-1}] = [L T^{-2}]$
Now, we can find the dimensions of force using the formula $F = ma$:
Dimension of Force = Dimension of Mass $\times$ Dimension of Acceleration
Dimension of Force = $[M] \times [L T^{-2}]$
Dimension of Force = $[M L T^{-2}]$
This is the dimensional formula for force. We can now compare this result with the given options:
- Option 1: $[M^{0} L^{3} T^{O}]$ - This represents the dimensions of volume.
- Option 2: $[M L^{-3} T^{0}]$ - This represents the dimensions of density (Mass/Volume).
- Option 3: $[M^{0} L T^{-1}]$ - This represents the dimensions of velocity (without mass dependence).
- Option 4: $[M L T^{-2}]$ - This matches our derived dimensions for force.
Therefore, the correct dimensions of force are $[M L T^{-2}]$.
Revision Table: Common Physical Quantity Dimensions
| Quantity | Formula | Dimensions |
|---|---|---|
| Length | - | $[L]$ |
| Mass | - | $[M]$ |
| Time | - | $[T]$ |
| Velocity | Displacement / Time | $[L T^{-1}]$ |
| Acceleration | Velocity / Time | $[L T^{-2}]$ |
| Force | Mass $\times$ Acceleration | $[M L T^{-2}]$ |
| Work / Energy | Force $\times$ Displacement | $[M L T^{-2} \times L] = [M L^2 T^{-2}]$ |
| Power | Work / Time | $[M L^2 T^{-2} / T] = [M L^2 T^{-3}]$ |
| Pressure | Force / Area | $[M L T^{-2} / L^2] = [M L^{-1} T^{-2}]$ |
Additional Information on Dimensional Analysis
Dimensional analysis is crucial in physics for several reasons:
- Checking Consistency: It can be used to check the dimensional consistency of equations. If an equation is physically correct, the dimensions on both sides must be the same.
- Deriving Relationships: In some cases, it can help in deriving relationships between physical quantities, especially in finding how one quantity depends on others.
- Unit Conversion: Understanding dimensions aids in converting units from one system (like SI) to another.
The dimensions $[M]$, $[L]$, and $[T]$ are considered fundamental dimensions in mechanics. Other quantities are derived from these. The representation of a physical quantity in terms of these fundamental dimensions raised to certain powers is called its dimensional formula or dimensional expression.
Important Questions from Dimensional formulae and dimensional equations
LT-2 is the dimension of which of the following quantities?
- Considering the Lorentz force $\vec{F} = q(\vec{v} \times \vec{B})$, where $F$ is force, $q$ is electric charge, and $v$ is velocity, what is the dimensional formula for magnetic flux density $B$?
- The characteristic impedance of free space, $Z_0$, is given by the expression $Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}}$. If $\mu_0$ represents the magnetic permeability and $\epsilon_0$ represents the electric permittivity, what are the dimensions of $Z_0$?
If \(x = a^3 + bt + ct^2\), where \(x\) is metres and \(t\) in seconds, then the unit of \(a\) is
- The dimensional formula of speed
